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Minimize L = 4x + 4y + zSubject tox + y + z ≤ 22x + y ≤ 3x ≥ 0, ≥ 0, z ≥ 0 |
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Answer» Given linear programming is objective function is Minimize L = 4x + 4y + z subject to x + y + z \(\leq\) 2 2x + y \(\leq\) 3 x \(\geq\)0, y \(\geq\) 0, z \(\geq\) 0 Converts given minimization problem into maximize problem: (Put -x in place of x, -y in place of y & -z in place of z in objective function) Then objective function is Maximize - L = -4x - 4y - z subject to x + y + z \(\leq\) 2 2x + y \(\leq\) 3 x \(\geq\) 0, y \(\geq\) 0, z \(\geq\) 0 Converts inequalities into equations, we get x + y + z + x1 + 0x2 = 2 2x + y + 0z + 0x1 + x2 = 3 x, y, z, x1, x2 \(\geq\) 0 First simplex table
\(\because\) All Zj - Cj \(\geq\) 0 And x, y, z are not present in basic variable \(\therefore\) x = 0, y = 0, z = 0 is the solution of redical maximize linear/programming problem. \(\therefore\) Max(-L) = -4 x 0 - 4 x 0 - 0 = 0 \(\therefore\) Min L = Max(-L) = 0 Hence, given linear programming is minimized at x = 0, y = 0, z = 0 and minimum value of objective function be 0. |
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